On showing that the usual basis $(e_n)_{n \in \Bbb N}$ is not Schauder for $\ell^\infty.$

Question

I am aware that there is more than one post about this in the forum already. Altought, I did a proof by myself and I have some doubts midway!

Too keep in mind: here, I am trying to prove that the basis $(e_n)_{n \in \Bbb N}$, where $e_n = (\delta_{nk})_{k \in \mathbb N}$ is not a Schauder basis for $\ell^\infty.$

Assume that $(e_n)_{n \in \mathbb N}$ is a Schauder basis for $\ell^\infty.$ Then, every sequence $x \in \ell^\infty$ admits a sequence of scalars $(\alpha_k)_{k \in \mathbb N} \subset \mathbb K$ (here, $\mathbb K \in \{\mathbb R,\mathbb C\}$) such that $$ \lim_{N\rightarrow \infty} \Bigg\| x -\sum_{k=1}^N\alpha_ke_k \Bigg\|= 0.$$ In particular, the sequence $x = (x_k)_{k \in \mathbb N} = (1,1,1,\dots)$ is in $\ell^\infty$ and thus, there exists a sequence of scalars $(\alpha_k)_{k \in \mathbb N} \subset \mathbb K$ such that $$ \lim_{N\rightarrow \infty} \Bigg\| (1,1,1,\dots) - \sum_{k=1}^N \alpha_ke_k\Bigg\| = 0.$$ But it is also true that $$ (1,1,1,\dots) - \sum_{k=1}^N \alpha_ke_k = (1-\alpha_1,1-\alpha_2,\dots,1-\alpha_N,1,1,1,\dots).$$ Thus, going back to the limit, we can say $$ \lim_{N\rightarrow \infty} \Bigg\| (1,1,1,\dots) - \sum_{k=1}^N \alpha_ke_k\Bigg\| = \lim_{N \rightarrow \infty} \Bigg\|\underbrace{(1-\alpha_1,1-\alpha_2,\dots,1-\alpha_N,1,1,1,\dots)}_{= (y_k)_{k \in \mathbb N}}\Bigg\| = \lim_{N \rightarrow \infty}\left( \sup_{k \in \Bbb N} |y_k| \right)$$ Now, it is obvious from how $(y_k)_{k \in \Bbb N}$ is defined that $\sup_{k \in \Bbb N} |y_k| \geqslant 1.$

I think I am almost done with the proof. If I can conclude that $$ 1 \leqslant \sup_{k \in \Bbb N} |y_k| \Rightarrow \lim_{N\rightarrow \infty} 1 \leqslant \lim_{N \rightarrow \infty} \left( \sup_{k\in \Bbb N} |y_k| \right)$$ The proof would be finished because then $1 \leqslant \lim_{N \rightarrow \infty} \left( \sup_{k\in \Bbb N} |y_k| \right) $ which implies that $\lim_{N \rightarrow \infty} \left( \sup_{k\in \Bbb N} |y_k| \right) \neq 0.$

I am just not sure if this last step (putting limits to infinity in both sides) of the inequality is valid.

Thanks for any help in advance.

Answer 1

You are making it too complicated. $ \sup_{k \in \Bbb N} |y_k| \geq 1$ for each $N$ and hence, $ \sup_{k \in \Bbb N} |y_k| $ does not tend to $0$ as $N \to \infty$

Answer 2

It is valid and follows from the general lemma:

Let $(a_n)_n$ and $(b_n)_n$ be real sequences such that $a_n \to a$ and $b_n \to b$. Then $$a_n \le b_n, \forall n \in \Bbb{N} \implies a \le b.$$

Now, sequences defined as $$a_n=1, \qquad b_n = \left\|(1,1,1,\dots) - \sum_{k=1}^n \alpha_ke_k\right\|_\infty$$ by assumption satisfy $a_n \to 1$, $b_n \to 0$ and $a_n \le b_n, \forall n\in\Bbb{N}$ so it follows $1 \le 0$ which is a contradiction.